Hyperfine
https://en.wikipedia.org/wiki/Hyperfine_structure http://www.feynmanlectures.caltech.edu/III_12.html "The order of magnitude of the classical interaction between two magnets would be the product of the two magnetic moments divided by the cube of the distance between them. The distance between the electron and the proton in the hydrogen atom is, speaking roughly, one half an atomic radius, or 0.5 angstrom. It is, therefore, possible to make a crude estimate that the constant A should be about equal to the product of the two magnetic moments μe and μp divided by the cube of 1/2 angstrom." This implies that A should actually be negative when the magnetic moments are Negative Hyperfine See Negative Hyperfine page... http://arxiv.org/pdf/1408.1347.pdf (Si29 paper) This paper has two nuclei used, one with a +ve and one with a -ve gyromagnetic ratio (GMR). Given Feynman's logic that A is a product of magnetic moments of electron and nucleus, then the sign should change when the sign of the GMR changes. However, both nuclei are shown to have positive hyperfines in this paper and no minus sign is introduced in the formulae to rectify any issue. Steger2011 follows a convention that the Hamiltonian uses different signs for electron and nucleus and allows them both to have positive GMR's, while the hyperfine is left positive. As below: H=\gamma_e S_z\cdot B_0 -\gamma_N I_z\cdot B_0+A S_z\cdot I_z This can be easily translated into the consistent, conventional "E=-μB" form by allowing the electron GMR to be defined by its true negative value, giving: H=-\gamma_e S_z\cdot B_0-\gamma_N I_z\cdot B_0+A S_z\cdot I_z H=-(\mu_e +\mu_N)\cdot B_0+A S_z\cdot I_z Which removes any intrinsic need for a negative hyperfine here. It seems clear from these two sources that negative hyperfines are not necessary even when the nuclei involved have opposite signs in their GMR. It seems a bit odd to me that an electron should experience a shift in energy in the same direction due to two nuclei in the same spin state but with opposite signed GMR's, because this implies that the magnetic moment of each is in opposite directions and hence one would expect that the shift felt by the electron would be different. The only explanation I can think of now is that hyperfine interaction always favours spins that anti-align, regardless of the GMR's. This goes against Feynman's logic in describing the hyperfine as a product of magnetic moments, but even then in his hydrogen example the logic failed because it was a case of opposite signs and the hyperfine of hydrogen is experimentally found to be positive. Feynman notesFeynman Lecture on Hydrogen Hyperfine http://www.feynmanlectures.caltech.edu/III_12.html "(We originally took A as positive because the theory we spoke of says it should be, and experimentally it is indeed so.)" despite earlier noting "μ_p is the magnetic moment of the proton, which is about 1000 times smaller than μ_e, and has the opposite sign". Interestingly, the lecture goes on to show how the hyperfine ground state of hydrogen (E=-4A) has no magnetic moment and would pass through a Stern-Gerlach apparatus undeflected. The state is calculated as the singlet state: |IV\rangle=\dfrac{|+\,-\rangle-|-\,+\rangle}{\sqrt{2}} Singlet states are known to have zero magnetic moment, but when you consider the opposite signs of the GMR's then the reason behind this is no longer so obvious. Shouldn't anti-aligned particles with opposite GMR's actually give a net magnetic moment? (continued in Negative Hyperfine page) References Category:Spin Category:Quantum Category:Physics